在矩形ABCD中,对角线AC BD交与点O,OE平分角ADC交BE于E点,角BDE=15°求角DOC和角COE?

OE平分角ADC -->∠CDE=45
∠CDO=∠CDE-∠BDE=45-15=30
∠DOC=180-∠CDO-∠DCO=180-2*30=120
∠CDO=30 --> CD=2BC,AB=2AD
∠CDE=45 --> AE=AD=AB/2
所以,OE是等腰三角形AOB的中线,
所以,OE是等腰三角形AOB的角平分线 -->∠BOE=∠AOB/2
∠COE=∠COB+∠BOE
=∠COB+∠AOB/2
=∠COB+(180-∠COB)/2
=90+∠COB/2
=90+2∠CDO/2
=90+∠CDO
=90+30
=120